\(\int \frac {1}{\sqrt {-2+3 x^2+3 x^4}} \, dx\) [44]

Optimal result

Integrand size = 16, antiderivative size = 146 \[ \int \frac {1}{\sqrt {-2+3 x^2+3 x^4}} \, dx=\frac {\sqrt {\frac {4-\left (3-\sqrt {33}\right ) x^2}{4-\left (3+\sqrt {33}\right ) x^2}} \sqrt {-4+\left (3+\sqrt {33}\right ) x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{33} x}{\sqrt {-4+\left (3+\sqrt {33}\right ) x^2}}\right ),\frac {1}{22} \left (11+\sqrt {33}\right )\right )}{2 \sqrt {2} \sqrt [4]{33} \sqrt {\frac {1}{4-\left (3+\sqrt {33}\right ) x^2}} \sqrt {-2+3 x^2+3 x^4}} \]

[Out]

1/132*EllipticF(33^(1/4)*x*2^(1/2)/(-4+x^2*(3+33^(1/2)))^(1/2),1/22*(242+22*33^(1/2))^(1/2))*((4-x^2*(3-33^(1/
2)))/(4-x^2*(3+33^(1/2))))^(1/2)*(-4+x^2*(3+33^(1/2)))^(1/2)*33^(3/4)*2^(1/2)/(3*x^4+3*x^2-2)^(1/2)/(1/(4-x^2*
(3+33^(1/2))))^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1112} \[ \int \frac {1}{\sqrt {-2+3 x^2+3 x^4}} \, dx=\frac {\sqrt {\frac {4-\left (3-\sqrt {33}\right ) x^2}{4-\left (3+\sqrt {33}\right ) x^2}} \sqrt {\left (3+\sqrt {33}\right ) x^2-4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{33} x}{\sqrt {\left (3+\sqrt {33}\right ) x^2-4}}\right ),\frac {1}{22} \left (11+\sqrt {33}\right )\right )}{2 \sqrt {2} \sqrt [4]{33} \sqrt {\frac {1}{4-\left (3+\sqrt {33}\right ) x^2}} \sqrt {3 x^4+3 x^2-2}} \]

[In]

Int[1/Sqrt[-2 + 3*x^2 + 3*x^4],x]

[Out]

(Sqrt[(4 - (3 - Sqrt[33])*x^2)/(4 - (3 + Sqrt[33])*x^2)]*Sqrt[-4 + (3 + Sqrt[33])*x^2]*EllipticF[ArcSin[(Sqrt[
2]*33^(1/4)*x)/Sqrt[-4 + (3 + Sqrt[33])*x^2]], (11 + Sqrt[33])/22])/(2*Sqrt[2]*33^(1/4)*Sqrt[(4 - (3 + Sqrt[33
])*x^2)^(-1)]*Sqrt[-2 + 3*x^2 + 3*x^4])

Rule 1112

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*(Sqrt[(2*a + (b + q)*x^2)/q]/(2*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2*a + (b + q)
*x^2)]))*EllipticF[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q)]], (b + q)/(2*q)], x]] /; FreeQ[{a, b, c}, x] && Gt
Q[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\frac {4-\left (3-\sqrt {33}\right ) x^2}{4-\left (3+\sqrt {33}\right ) x^2}} \sqrt {-4+\left (3+\sqrt {33}\right ) x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{33} x}{\sqrt {-4+\left (3+\sqrt {33}\right ) x^2}}\right )|\frac {1}{22} \left (11+\sqrt {33}\right )\right )}{2 \sqrt {2} \sqrt [4]{33} \sqrt {\frac {1}{4-\left (3+\sqrt {33}\right ) x^2}} \sqrt {-2+3 x^2+3 x^4}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\sqrt {-2+3 x^2+3 x^4}} \, dx=-\frac {i \sqrt {4-6 x^2-6 x^4} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {6}{3+\sqrt {33}}} x\right ),-\frac {7}{4}-\frac {\sqrt {33}}{4}\right )}{\sqrt {-3+\sqrt {33}} \sqrt {-2+3 x^2+3 x^4}} \]

[In]

Integrate[1/Sqrt[-2 + 3*x^2 + 3*x^4],x]

[Out]

((-I)*Sqrt[4 - 6*x^2 - 6*x^4]*EllipticF[I*ArcSinh[Sqrt[6/(3 + Sqrt[33])]*x], -7/4 - Sqrt[33]/4])/(Sqrt[-3 + Sq
rt[33]]*Sqrt[-2 + 3*x^2 + 3*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.59 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {2 \sqrt {1-\left (\frac {3}{4}-\frac {\sqrt {33}}{4}\right ) x^{2}}\, \sqrt {1-\left (\frac {3}{4}+\frac {\sqrt {33}}{4}\right ) x^{2}}\, F\left (\frac {\sqrt {3-\sqrt {33}}\, x}{2}, \frac {i \sqrt {6}}{4}+\frac {i \sqrt {22}}{4}\right )}{\sqrt {3-\sqrt {33}}\, \sqrt {3 x^{4}+3 x^{2}-2}}\)	\(84\)
elliptic	\(\frac {2 \sqrt {1-\left (\frac {3}{4}-\frac {\sqrt {33}}{4}\right ) x^{2}}\, \sqrt {1-\left (\frac {3}{4}+\frac {\sqrt {33}}{4}\right ) x^{2}}\, F\left (\frac {\sqrt {3-\sqrt {33}}\, x}{2}, \frac {i \sqrt {6}}{4}+\frac {i \sqrt {22}}{4}\right )}{\sqrt {3-\sqrt {33}}\, \sqrt {3 x^{4}+3 x^{2}-2}}\)	\(84\)

[In]

int(1/(3*x^4+3*x^2-2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/(3-33^(1/2))^(1/2)*(1-(3/4-1/4*33^(1/2))*x^2)^(1/2)*(1-(3/4+1/4*33^(1/2))*x^2)^(1/2)/(3*x^4+3*x^2-2)^(1/2)*E
llipticF(1/2*(3-33^(1/2))^(1/2)*x,1/4*I*6^(1/2)+1/4*I*22^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.28 \[ \int \frac {1}{\sqrt {-2+3 x^2+3 x^4}} \, dx=-\frac {1}{24} \, {\left (\sqrt {33} \sqrt {-2} - 3 \, \sqrt {-2}\right )} \sqrt {\sqrt {33} + 3} F(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {33} + 3}\right )\,|\,\frac {1}{4} \, \sqrt {33} - \frac {7}{4}) \]

[In]

integrate(1/(3*x^4+3*x^2-2)^(1/2),x, algorithm="fricas")

[Out]

-1/24*(sqrt(33)*sqrt(-2) - 3*sqrt(-2))*sqrt(sqrt(33) + 3)*elliptic_f(arcsin(1/2*x*sqrt(sqrt(33) + 3)), 1/4*sqr
t(33) - 7/4)

Sympy [F]

\[ \int \frac {1}{\sqrt {-2+3 x^2+3 x^4}} \, dx=\int \frac {1}{\sqrt {3 x^{4} + 3 x^{2} - 2}}\, dx \]

[In]

integrate(1/(3*x**4+3*x**2-2)**(1/2),x)

[Out]

Integral(1/sqrt(3*x**4 + 3*x**2 - 2), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {-2+3 x^2+3 x^4}} \, dx=\int { \frac {1}{\sqrt {3 \, x^{4} + 3 \, x^{2} - 2}} \,d x } \]

[In]

integrate(1/(3*x^4+3*x^2-2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(3*x^4 + 3*x^2 - 2), x)

Giac [F]

\[ \int \frac {1}{\sqrt {-2+3 x^2+3 x^4}} \, dx=\int { \frac {1}{\sqrt {3 \, x^{4} + 3 \, x^{2} - 2}} \,d x } \]

[In]

integrate(1/(3*x^4+3*x^2-2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(3*x^4 + 3*x^2 - 2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {-2+3 x^2+3 x^4}} \, dx=\int \frac {1}{\sqrt {3\,x^4+3\,x^2-2}} \,d x \]

[In]

int(1/(3*x^2 + 3*x^4 - 2)^(1/2),x)

[Out]

int(1/(3*x^2 + 3*x^4 - 2)^(1/2), x)